Product of two CAT($kappa$) spaces is CAT($kappa$) for $kappa ge 0$
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I would like to see a "metric proof" that if two metric spaces $X$ and $Y$ are CAT($kappa$) for some $kappa ge 0$, then so is their product. I would be satisfied to see a proof for $X=Y=S^2$.
By "metric proof" I mean one which does not rely on Riemannian geometry, but rather only uses Alexandrov (metric) geometry. I already understand the case where $kappa le 0$ (where in fact the product will only be CAT($textmax(0,kappa)$).
metric-geometry
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up vote
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favorite
I would like to see a "metric proof" that if two metric spaces $X$ and $Y$ are CAT($kappa$) for some $kappa ge 0$, then so is their product. I would be satisfied to see a proof for $X=Y=S^2$.
By "metric proof" I mean one which does not rely on Riemannian geometry, but rather only uses Alexandrov (metric) geometry. I already understand the case where $kappa le 0$ (where in fact the product will only be CAT($textmax(0,kappa)$).
metric-geometry
What is $rm CAT[k]$ and their product ?
â HK Lee
Sep 6 at 11:36
A CAT($kappa$) space is a geodesic space all of whose geodesic triangles of perimeter less than $2D_kappa$ satisfy the CAT$(kappa)$ inequality. Here $D_kappa= +infty$ if $kappa leq 0$ or $fracpikappa$ if $kappa > 0$. One version of the CAT$(kappa$) inequality says that if one takes such a triangle in $X$ and a comparison triangle in the model plane of curvature $kappa$, then the Alexandrov angles in the triangle in $X$ are no greater than the corresponding angles of the comparison triangle.
â Delfador Logalmier
Sep 6 at 12:17
The product of two metric spaces $Xtimes Y$ has the usual cartesian product $X times Y$ as its underlying set, and the metric is defined in the expected way: $$ d_X times Y((x_1,y_1),(x_2,y_2))^2 = d_X(x_1,x_2)^2+d_Y(y_1,y_2)^2 $$
â Delfador Logalmier
Sep 6 at 12:18
Sorry, in the first comment, $ fracpikappa$ should have been $fracpisqrtkappa$. $X$ is the CAT$(kappa)$ space under discussion.
â Delfador Logalmier
Sep 6 at 12:24
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I would like to see a "metric proof" that if two metric spaces $X$ and $Y$ are CAT($kappa$) for some $kappa ge 0$, then so is their product. I would be satisfied to see a proof for $X=Y=S^2$.
By "metric proof" I mean one which does not rely on Riemannian geometry, but rather only uses Alexandrov (metric) geometry. I already understand the case where $kappa le 0$ (where in fact the product will only be CAT($textmax(0,kappa)$).
metric-geometry
I would like to see a "metric proof" that if two metric spaces $X$ and $Y$ are CAT($kappa$) for some $kappa ge 0$, then so is their product. I would be satisfied to see a proof for $X=Y=S^2$.
By "metric proof" I mean one which does not rely on Riemannian geometry, but rather only uses Alexandrov (metric) geometry. I already understand the case where $kappa le 0$ (where in fact the product will only be CAT($textmax(0,kappa)$).
metric-geometry
metric-geometry
asked Sep 6 at 11:04
Delfador Logalmier
112
112
What is $rm CAT[k]$ and their product ?
â HK Lee
Sep 6 at 11:36
A CAT($kappa$) space is a geodesic space all of whose geodesic triangles of perimeter less than $2D_kappa$ satisfy the CAT$(kappa)$ inequality. Here $D_kappa= +infty$ if $kappa leq 0$ or $fracpikappa$ if $kappa > 0$. One version of the CAT$(kappa$) inequality says that if one takes such a triangle in $X$ and a comparison triangle in the model plane of curvature $kappa$, then the Alexandrov angles in the triangle in $X$ are no greater than the corresponding angles of the comparison triangle.
â Delfador Logalmier
Sep 6 at 12:17
The product of two metric spaces $Xtimes Y$ has the usual cartesian product $X times Y$ as its underlying set, and the metric is defined in the expected way: $$ d_X times Y((x_1,y_1),(x_2,y_2))^2 = d_X(x_1,x_2)^2+d_Y(y_1,y_2)^2 $$
â Delfador Logalmier
Sep 6 at 12:18
Sorry, in the first comment, $ fracpikappa$ should have been $fracpisqrtkappa$. $X$ is the CAT$(kappa)$ space under discussion.
â Delfador Logalmier
Sep 6 at 12:24
add a comment |Â
What is $rm CAT[k]$ and their product ?
â HK Lee
Sep 6 at 11:36
A CAT($kappa$) space is a geodesic space all of whose geodesic triangles of perimeter less than $2D_kappa$ satisfy the CAT$(kappa)$ inequality. Here $D_kappa= +infty$ if $kappa leq 0$ or $fracpikappa$ if $kappa > 0$. One version of the CAT$(kappa$) inequality says that if one takes such a triangle in $X$ and a comparison triangle in the model plane of curvature $kappa$, then the Alexandrov angles in the triangle in $X$ are no greater than the corresponding angles of the comparison triangle.
â Delfador Logalmier
Sep 6 at 12:17
The product of two metric spaces $Xtimes Y$ has the usual cartesian product $X times Y$ as its underlying set, and the metric is defined in the expected way: $$ d_X times Y((x_1,y_1),(x_2,y_2))^2 = d_X(x_1,x_2)^2+d_Y(y_1,y_2)^2 $$
â Delfador Logalmier
Sep 6 at 12:18
Sorry, in the first comment, $ fracpikappa$ should have been $fracpisqrtkappa$. $X$ is the CAT$(kappa)$ space under discussion.
â Delfador Logalmier
Sep 6 at 12:24
What is $rm CAT[k]$ and their product ?
â HK Lee
Sep 6 at 11:36
What is $rm CAT[k]$ and their product ?
â HK Lee
Sep 6 at 11:36
A CAT($kappa$) space is a geodesic space all of whose geodesic triangles of perimeter less than $2D_kappa$ satisfy the CAT$(kappa)$ inequality. Here $D_kappa= +infty$ if $kappa leq 0$ or $fracpikappa$ if $kappa > 0$. One version of the CAT$(kappa$) inequality says that if one takes such a triangle in $X$ and a comparison triangle in the model plane of curvature $kappa$, then the Alexandrov angles in the triangle in $X$ are no greater than the corresponding angles of the comparison triangle.
â Delfador Logalmier
Sep 6 at 12:17
A CAT($kappa$) space is a geodesic space all of whose geodesic triangles of perimeter less than $2D_kappa$ satisfy the CAT$(kappa)$ inequality. Here $D_kappa= +infty$ if $kappa leq 0$ or $fracpikappa$ if $kappa > 0$. One version of the CAT$(kappa$) inequality says that if one takes such a triangle in $X$ and a comparison triangle in the model plane of curvature $kappa$, then the Alexandrov angles in the triangle in $X$ are no greater than the corresponding angles of the comparison triangle.
â Delfador Logalmier
Sep 6 at 12:17
The product of two metric spaces $Xtimes Y$ has the usual cartesian product $X times Y$ as its underlying set, and the metric is defined in the expected way: $$ d_X times Y((x_1,y_1),(x_2,y_2))^2 = d_X(x_1,x_2)^2+d_Y(y_1,y_2)^2 $$
â Delfador Logalmier
Sep 6 at 12:18
The product of two metric spaces $Xtimes Y$ has the usual cartesian product $X times Y$ as its underlying set, and the metric is defined in the expected way: $$ d_X times Y((x_1,y_1),(x_2,y_2))^2 = d_X(x_1,x_2)^2+d_Y(y_1,y_2)^2 $$
â Delfador Logalmier
Sep 6 at 12:18
Sorry, in the first comment, $ fracpikappa$ should have been $fracpisqrtkappa$. $X$ is the CAT$(kappa)$ space under discussion.
â Delfador Logalmier
Sep 6 at 12:24
Sorry, in the first comment, $ fracpikappa$ should have been $fracpisqrtkappa$. $X$ is the CAT$(kappa)$ space under discussion.
â Delfador Logalmier
Sep 6 at 12:24
add a comment |Â
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What is $rm CAT[k]$ and their product ?
â HK Lee
Sep 6 at 11:36
A CAT($kappa$) space is a geodesic space all of whose geodesic triangles of perimeter less than $2D_kappa$ satisfy the CAT$(kappa)$ inequality. Here $D_kappa= +infty$ if $kappa leq 0$ or $fracpikappa$ if $kappa > 0$. One version of the CAT$(kappa$) inequality says that if one takes such a triangle in $X$ and a comparison triangle in the model plane of curvature $kappa$, then the Alexandrov angles in the triangle in $X$ are no greater than the corresponding angles of the comparison triangle.
â Delfador Logalmier
Sep 6 at 12:17
The product of two metric spaces $Xtimes Y$ has the usual cartesian product $X times Y$ as its underlying set, and the metric is defined in the expected way: $$ d_X times Y((x_1,y_1),(x_2,y_2))^2 = d_X(x_1,x_2)^2+d_Y(y_1,y_2)^2 $$
â Delfador Logalmier
Sep 6 at 12:18
Sorry, in the first comment, $ fracpikappa$ should have been $fracpisqrtkappa$. $X$ is the CAT$(kappa)$ space under discussion.
â Delfador Logalmier
Sep 6 at 12:24